????^3+????^2−4????=4 How to factorise?
回答 (3)
2016-02-28 2:09 pm
x³ + x² - 4x = 4
x³ + x² - 4x - 4 = 0
(x³ + x²) - (4x + 4) = 0
x²(x + 1) - 4(x + 1) = 0
(x + 1)(x² - 4) = 0
(x + 1)(x² - 2²) = 0
(x + 1)(x + 2)(x - 2) = 0
x + 1 = 0 or x + 2 = 0 or x - 2 = 0
x = -1 or x = -2 or x = 2
x³ + x² - 4x - 4 = 0
(x³ + x²) - (4x + 4) = 0
x²(x + 1) - 4(x + 1) = 0
(x + 1)(x² - 4) = 0
(x + 1)(x² - 2²) = 0
(x + 1)(x + 2)(x - 2) = 0
x + 1 = 0 or x + 2 = 0 or x - 2 = 0
x = -1 or x = -2 or x = 2
2016-02-28 2:14 pm
x^3 + x^2 - 4x - 4 = 0;
(x^2 - 4)(x + 1) = 0 =>
(x+2) (x-2) (x+1) = 0.
(x^2 - 4)(x + 1) = 0 =>
(x+2) (x-2) (x+1) = 0.
2016-02-28 2:03 pm
Do you mean how to factor ????^3+????^2−4???? - 4 ? That would be (x+1) ( x-2) (x + 2)
收錄日期: 2021-04-18 14:31:30
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